ok but I am not sure where you want me to start.
I will assume for starters that you accept my diagram.
Since I said to let the sides of all the little euqilatoreral triangles be 2x, The height of the little tianges will be
\(h=\sqrt{(2x)^2-x^2}\\ h=\sqrt{3x^2}\\ h=\sqrt{3}\;x\\ Area\; of \;the \;little \;triangles\; = \sqrt3\; x^2 \)
Area of the sector ADD'
\(=\frac{60}{360}*\pi * (2x)^2\\ =\frac{\pi}{6}*4x^2\\ =\frac{2\pi x^2}{3}\)
Area of the little segment
\(=\frac{2\pi x^2}{3}-\sqrt3\;x^2\\ =\frac{2\pi x^2-3\sqrt3\;x^2}{3}\\ =\frac{x^2}{3}(2\pi -3\sqrt3)\\\)
The area of the triangle outside the circle
\(=\frac{3\sqrt3 x^2 }{3}-\frac{2\pi x^2-3\sqrt3 x^2}{3}\\ =\frac{3\sqrt3 x^2 -2\pi x^2+3\sqrt3 x^2 }{3}\\ =\frac{ x^2}{3}*(6\sqrt3 -2\pi )\\\)
But we are told that this area is 1
\(\frac{ x^2}{3}*(6\sqrt3 -2\pi )=1\\ x^2=\frac{-3}{2\pi-6\sqrt3}\\ x=\sqrt{\frac{-3}{2\pi-6\sqrt3}}\\ AB=4x\\ AB=4\sqrt{\frac{-3}{2\pi-6\sqrt3}}\approx 3.4178 \quad (edited: \;due \;to\; a \;copy \; error \;41 \;not \;14) \)
This answer looks really awfull so it is quite likely that I made a careless error somewhere.
You need to check my working and logic.
LaTex:
h=\sqrt{(2x)^2-x^2}\\
h=\sqrt{3x^2}\\
h=\sqrt{3}\;x\\
Area\; of \;the \;little \;triangles\; = \sqrt3\; x^2
=\frac{60}{360}*\pi * (2x)^2\\
=\frac{\pi}{6}*4x^2\\
=\frac{2\pi x^2}{3}
=\frac{2\pi x^2}{3}-\sqrt3\;x^2\\
=\frac{2\pi x^2-3\sqrt3\;x^2}{3}\\
=\frac{x^2}{3}(2\pi -3\sqrt3)\\
=\frac{3\sqrt3 x^2 }{3}-\frac{2\pi x^2-3\sqrt3 x^2}{3}\\
=\frac{3\sqrt3 x^2 -2\pi x^2+3\sqrt3 x^2 }{3}\\
=\frac{ x^2}{3}*(6\sqrt3 -2\pi )\\
\frac{ x^2}{3}*(6\sqrt3 -2\pi )=1\\
x^2=\frac{-3}{2\pi-6\sqrt3}\\
x=\sqrt{\frac{-3}{2\pi-6\sqrt3}}\\
AB=4x\\
AB=4\sqrt{\frac{-3}{2\pi-6\sqrt3}}\approx 3.4178 (edited \;due \;to\; a \;copy\; error)
